[ Mektubu göndermeye gerek kalmadı çünkü kendim çözdüm.] Dear Prof. Ducheyne,

I'm reading your paper The Cavendish Experiment as a Tool for Historical Understanding of Science to better understand Cavendish's original analysis.

First of all, thanks for the detailed explanations that I could not find in other similar publications studying Cavendish experiment.

I would like to ask you a question about Cavendish's comparison of his torsion pendulum with a seconds pendulum.

If I understand correctly, he first assumes that his torsion pendulum oscillates as if under a restoring force which is equivalent to \(g\), the acceleration due to gravity. With this assumption he can connect the properties of the vertical pendulum with his torsion pendulum.

Then from the pendulum law, as you explain in your footnote 22, he finds,

\begin{equation*} t_1 :: \sqrt{\frac{x_1}{x_2}} = \sqrt{\frac{36.65}{39.14}} \end{equation*}

Then, he wants to find the period for any stiffnes of the wire. Cavendish writes, "…if the stiffness of the wire is such as to make it vibrate in \(N\) seconds, the force which must be applied to each ball, in order to draw it aside by an angle \(A\), is to the weight of the ball as the arch of \(A \times 1/N^2 \times 36.65/39.14\) to the radius."

I don't understand what Cavendish does here. It seems to me that the force required to pull the arm aside is only proportional to the stiffness of the wire, not to the weight of the ball.

What confuses me is Cavendish's use of the two forces, the restoring force of the wire and the force "that must be applied to each ball."

Your explanation at this point is helpful but I still did not understand what's going on.

You wrote, "Here Cavendish pointed out that the force exerted on the balls (\(F_e\)) swinging along a pendulum is to the restoring force (\(F_r\)) as \(T^2/N^2\), so

\begin{equation*} F_r :: \frac{T^2}{N^2} \end{equation*}

Here, \(T^2 = \frac{36.65}{39.14}\).

Where does Cavendish point this out? Why is the ratio of the lengths of the torsion pendulum and the seconds pendulum proportional to the restoring force?

You continue:

Because the restoring force is, furthermore proportional to the weight of the ball (\(W_b\)) times the arch of A…"

So, \(F_r :: W_b \times\) arch of A

Why is this so? I don't understand.

You continue:

"…it follows that,

\begin{align*} \frac{F_e}{W_b} &:: \text{arch of} A \times \frac{T^2}{N^2}\\ \\ &= \text{arch of} A \times \frac{36.65}{39.14} \times \frac{1}{N^2} \end{align*}

So how does Cavendish arrive at the period for a restoring force, for a wire of any stiffness?

Any more explanations you may have on these questions will be very welcome.

Cavendish also relates the weight of the ball \((W_b)\) on the pendulum with the force (\(F_e\)) required to move the pendulum arm. But, it seems to me that, the force that must be applied to the ball is not proportional to the weight of the ball. The symmetry of the torsion pendulum makes the balls weightless, that's why it becomes possible to measure extremely small forces. On the Cavendish pendulum, given the same stiffness for the wire, a ball of 1 kg and a ball of 100 kg will be attracted by the same force. Is this correct?

Once I thoroughly understand Cavendish's analysis, I'm hoping to replicate the experiment according to Cavendish's exact specifications.

Many thanks and best wishes.